hash entry

Only discuss the most common binary hash.

example

Luo Gu P3370

Meaning of the questions: How many seek different string ($ N \ leq 10 ^ 4 $) N string there.

analysis:

Seeking a hash value for each string, how many different hash values ​​are then statistically there.

Single Hash

Select a large prime number, or natural overflow.

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
const ull base = 233;
ull mod=212370440130137957ll;
const int maxn = 10000+10;
int n;
wool f [maxn];
char s[maxn];

the Hash ULL ( char S [])   // Nature spill 
{
    wool ans = 0 ;
    int len = strlen (S);
    for ( int i = 0 ; i <len; i ++ )
    {
        ans = ( base * + ANS (ULL) s [i])% v;
    }
    return years;
}

set<ull>st;
int main()
{
    scanf("%d", &n);
    for(int i = 0;i < n;i++)
    {
        scanf("%s", s);
        st.insert(Hash(s));
    }
    printf("%d\n", st.size());
}

Double hashing

The use of two modular, hash collision probability will be much lower, but constant increases.

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
const ull base = 233;
wool mod1 = 212370440130137957ll;
ull mod2 = 1 << 30;
const int maxn = 10000+10;
int n;
char s[maxn];

Hashl ULL ( char S [])   // Nature spill 
{
    wool ans = 0 ;
    int len = strlen (S);
    for ( int i = 0 ; i <len; i ++ )
    {
        years = ( base * year + (ull) s [i])% mod1;
    }
    return years;
}
Hash2 ULL ( char S [])   // Nature spill 
{
    wool ans = 0 ;
    int len = strlen (S);
    for ( int i = 0 ; i <len; i ++ )
    {
        years = ( base * year + (ull) s [i])% mod2;
    }
    return years;
}

struct Node{
    wool x, y;
}f[maxn];
bool cmp(Node a, Node b)
{
    if(a.x == b.x)  return a.y < b.y;
    return a.x < b.x;
}

int main ()
{
    scanf("%d", &n);
    for(int i = 0;i < n;i++)
    {
        scanf("%s", s);
        f[i].x = Hash1(s);
        f[i].y = Hash2(s);
        //printf("%lld %lld\n", f[i].x, f[i].y);
    }
    sort(f, f+n, cmp);
    int cnt  = 1;
    for(int i = 0;i < n-1;i++)
    {
        if(f[i+1].x != f[i].x || f[i+1].y != f[i].y)  cnt++;
    }
    printf("%d\n", cnt);
}

 

 

Reference Links: https://www.cnblogs.com/henry-1202/p/8868414.html